Extracting Square Roots with pencil and paper

Welcome and have fun! To ensure, the HAFAL-factor (have-fun-and-learn) is maximum, you should be familiar with performing of simple multiplication (like e.g. 9x4809) and division (like e.g. 659:34) with pencil and paper. If so, you won't have any problems to extract square roots by pencil and paper! Let's have some fun together...

We will see an example including detailed explanations. It will be a standard calculation, not having any exeptional case.
Another example is fully loaded with 'special cases' that - I felt -should have been mentioned for proper understanding.
And finally we'll have the opportunity to practice our new knowledge on some - already calculated - examples. But enough theory for the moment. Let's do some fun exercise.We are going to extract the square root of 283.024. And believe me, it's not the most difficult thing on planet earth...

Before we start, already a quick look at the final result of the calculation in the following figure A. Doesn't look that odd, hm? Almost like a standard division...

figure 0: Start of calculation

Write down the number 283.024 usually under a square root sign and separate the number by vertical lines into groups, containing each two digits. Start counting the groups from the right-hand side. See from figure 0 how it looks like.

Note:
If there were a comma (e.g. 2830,24) you'd have to start drawing the first line directly below the comma and separate the number from there, counting groups to the left as well as to the right-hand side.

figure 1: How it looks after Step1

Start with the very left group (group 1) of the number, which is 28.
Look for the number that, when you square it (x²), is nearest but smaller than, or exactly 28. The answer is 5, because 5² = 25, and 25 is the nearest square but smaller than 28. The next x² would have been 6² = 36, which is already bigger than 28.
Therfore, 5 is the first digit of the solution!

Write now 5 to the right-hand side of the "=" sign and 25 below 28 into line 1 (L1).

figure 2: How it looks after Step2

Subtract 25 from 28 (L0-L1), which is 3 (the 'remainder'), and write this 3 below 25 (into L2). Get the left digit of the next group (group 2), which is 3, and write it beside the 3 in L2, so that finally stands 33 there.

Perform 'intermediate division' to prepare getting the next digit of the solution. I.e., divide the value in L2, which is 33, by 2 x the current 'result' of the root, which is 2x5, means 33:(2x5). The result of this is 3,3 or in other words 3 + a rest. The decimals (the 'rest') are not of any interest in this calculation. The 3 from the left handside of the comma is the result of this Step 2 and will be used in the next step (Step 2a).

figure 2a: How it looks after Step 2a

Perform 'intermediate multiplication' like described in figure 2a.

I.e., multiply the result of previously performed 'intermediate division' (Step 2), which is 3, by the number assembled by "2 x the current 'result' of the root", which is 2x5, and the result of Step 2, which is 3. So, 3x(10 & 3) = 3x103 = 309.

figure 2b: How it looks after Step 2b

Get now the right digit of the next group (group 2), which is 0, and write it beside the 33 in L2, so that finally stands 330 there.

Compare the result of previously performed 'intermediate multiplication' (Step 2a), which is 309, with the number in L2, which is now 330 and check that the result of Step 2a is either smaller than or equal to the number in L2.
Yes, 309 is smaller than 330.
Therefore 3 is the next digit of the solution!

Note:
If the result of Step 2a were bigger than the number in L2, it would have been a special case. This will be explained later.

figure 3: How it looks after Step3

Subtract now 309 from 330 (L2-L3), which is 21 (the 'remainder'), and write this 21 below the 309 (into L4). Get the left digit of the next group (group 3), which is 2, and write it beside the 21 in L4, so that finally stands 212 there.

Perform 'intermediate division' (like in Step 2) to prepare getting the next digit of the solution. I.e., divide the value in L4, which is 212, by 2 x the current 'result' of the root, which is 2x53, means 212:(2x53). The result is exactly 2. This 2 is the result of this Step 3 and will be used in the next step (Step 3a).

Note:
If there were decimals in the result of this step, they would not have been of interest here.

figure 3a: How it looks after Step3a

Perform 'intermediate multiplication' like described in figure 3a.

I.e., multiply the result of previously performed 'intermediate division' (Step 3), which is 2, by the number assembled by "2 x the current 'result' of the root", which is 2x53 and the result of Step 3, which is 2. So, 2x(106 & 2) = 2x1062 = 2124.

figure 3b: How it looks after Step3b

Get now the right digit of the next group (group 3), which is 4, and write it beside the 212 in L4, so that finally stands 2124 there.

Compare the result of previously performed 'intermediate multiplication' (Step 3a), which is 2124, with the number in L4, which is also 2124 and check that the result of Step 3a is either smaller than or equal to the number in L4. Yes, 2124 is equal to 2124.
Therefore 2 is the next digit of the solution!

Note:
If the result of Step 3a were bigger than the number in L4, it would have been a special case. This will be explained later.

figure 4: How it looks after Step4

Subtract 2124 from 2124 (L4-L5), which is 0 (the 'remainder'), and write this 0 below the 2124 (into L6).

Now, there is no next group anymore below the square root sign (rightmost group already reached) and the result of the last performed subtraction (in L6) is 0.
This combination ("in rightmost group" & "remainer = 0") means, the exact solution for the calculation was found!

When the rightmost group is reached but the 'remainder' is still bigger than zero, the exact solution of the problem was not yet found. The calculation can be continued by adding more groups to the right-hand side of the number. Each of those groups contains '00' as value then. The more groups are added (i.e., the longer calculating will be continued) the more accurate the solution will be.

As you maybe noticed, there is a periodical way of reaching the solution:

Start!
Looking for a 'square' (x²),
Division, Multiplication, Subtraction,
Division, Multiplication, Subtraction,
Division, Multiplication, Subtraction,
...
Division, Multiplication, Subtraction,
Solution!

figure 5: Special Case 1

Special Case 1:
'intermediate division' containes more than one digit.

Result of 'intermediate division', which is 11, contains two (more than one!) digits. But it must contain only one digit! Therefore reduce 11 to the next lower 1-digit-number, which is 9.
In the next step perform 'intermediate multiplication' with the new result, which is now 9.

Rule: Any 2-digit-number as result of 'intermediate division' to be reduced to 9.

figure 6: Special Case 2

Special Case 2:
'intermediate multiplication': (Left > Right)!

Perform 'intermediate multiplication' (see from step 2a, figure 2a.
Means here, 9 x (2 & 9) = 9x29 = 261. This value is now bigger than the value in L2, which is 225. But it must be smaller than or equal to the value in L2, not bigger!
Therefore reduce the digit 9 to one digit lower, which is 8.
Perform 'intermediate multiplication' again, but now with the new digit 8. The result is 224, which is smaller than 225 (L2). Now, it can be continued calculating as usual.

Note:
Reduction of the digits can theoretically go down until zero. Try e.g. square root of 4,206601=2,051.

Rule: If "Left" > "Right" in 'intermediate multiplication', reduce digit one step (digit = digit - 1) and perform 'intermediate multiplication' again.

figure 7: Special Case 3

Special Case 3:
'intermediate division' is zero ('0').

Result of 'intermediate division' is zero ('0'). This is actually not a special case, but it it may look like one and could confuse. Therefore it was mentioned here.

The result of 'intermediate multiplication' says 0 < 180. This is correct and therefore 0 is the next digit of the solution.

Rule: If result of 'intermediate division' is 0, continue as usual (append it as usual to the current result of the root). No exeption.

figure 8: Special Case 4

Special Case 4:
'intermediate division': Ignore comma in current result of the root.

First, don't forget to set the comma after the 18 (current result of the root).
Perform 'intermediate division' (like in Step 2) to prepare getting the next digit of the solution. I.e., divide the value in L6, which is 18025, by 2 x the current 'result' of the root, which is 2x180, means 212:(2x180), NOT 212:(2x18,0)!! This is important to remember. If it isn't handled like this, the result will be wrong!

Rule: If there is a comma in the current result of the root, ignore it for 'intermediate division'.

figure 9: Special Case 5

Special Case 5:
Division by zero!

Rule: If in the very beginning of calculation a group of two digits contains only zero(s), a zero for each group shall be appended to the current result of the root, without calculating. After this continue calculating as usual.

I hope you had fun and could follow the explanations. If you have further questions, improvement proposals or you just want to tell me something ... feel absolutely free to email me to achim_behr@hotmail.com. If you'd like to create a link to this site on your homepage, just do so. I'd be pleased.

Usually a 'solution' of a square root can easily be given in some couples of minutes with let's say 5 to 6 digits accuracy - using this method. Higher accuracy requires definitely more time and patience, due to increasing numbers in the intermedate calculations. If you should have enough spare time to share, don't care about the previous sentence. :-)

You can also extract 'manually' square roots by using Napier rods. It's faster and there are fewer opportunities to make mistakes. Interested? - Or you'd maybe just like to know what a Napier rod is? Look here.

You can do the same also with a Otis King Pocket Calculator or a Slide Rule, but not as accurate as with Napier rods or pencil & paper, ...but slightly faster.

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